Question

A cold duck slides down a snow covered hill inclined at an angle of 35° to the horizontal. If the ducks speed is a constant 5m/s, determine the horizontal and vertical components of its velocity.

Answer 1

Given data:

* The angle of the inclined plane is,

`\theta=35^(\circ)`

* The speed of the duck along the inclined plane is v = 5 m/s.

Solution:

The diagrammatic representation of the given case is,

From the triangle ABO,

`\angle ABO=180^(\circ)-\angle AOB-\angle OAB`

Substituting the known values,

`\begin{gathered} \angle ABO=180^(\circ)-90^(\circ)-35^(\circ) \\ \angle ABO=55^(\circ) \end{gathered}`

Thus, the inclined plane makes an angle of 55 degrees with the negative of the y-axis.

Similarly, the angle of the velocity of the duck along the inclined plane with the negative y-axis is 55 degrees.

Thus, the vertical component of the velocity is,

`\begin{gathered} v_y=-v\cos (55^(\circ)) \\ v_y=-5*\cos (55^(\circ)) \\ v_y=-2.88\text{ m/s} \end{gathered}`

The horizontal component of the velocity is,

`\begin{gathered} v_x=v\sin (55^(\circ)) \\ v_x=5\sin (55^(\circ)) \\ v_x=4.096\text{ m/s} \end{gathered}`

Thus, the horizontal component of velocity is 4.096 meters per second, and the vertical component of velocity is -2.88 meters per second.